All categories

2 years ago

Environmental Effects Summary

1 answer:

5 0

River samples may evidence problems with the water treatment plant because it may contain an excessive amount of certain substances which are filtered during this procedure.

<h3>What is a water treatment plant?</h3>

A water treatment plant is a place where wastewater from a city is processed and purified in order to obtain potable water.

Water treatment procedure generally involves the collection, chemical substance addition, filtration, and disinfection.

Any type of problem in these steps can be evidenced by taking a sample of the water after treatment.

In conclusion, river samples evidence problems due to an excessive amount of certain substances which are filtered during this procedure.

Learn more about water treatment here:

brainly.com/question/13348717

#SPJ1

You might be interested in

How did you know the oxygen became a gas?

wel

Explain the question more please hggghhg

5 0

3 years ago

Read 2 more answers

Hippuric acid (HC9H8NO3)(HC9H8NO3), found in horse urine, has pKa=3.62pKa=3.62. Part A Calculate the pHpH in 0.140 MM hippuric a

Oxana [17]

Answer:

The pH in 0.140 M hippuric acid solution is 2.2.

Explanation :

Dissociation constant of the acid = $K_a$

$pK_a=-\log[K_a]$

$3.62=-\log[K_a]$

$K_a=2.399\times 10^{-4}$

Concentration of hippuric acid = c = 0.140 M

$HC_9H_8NO_3\rightleftharpoons C_9H_8NO_{3}^-+H^+$

Initially

c 0 0

At equilibrium

(c-x) x x

Concentration of acid = c $[HC_9H_8NO_3]=0.140 M$

Dissociation constant of an acid is given by:

$K_a=\frac{[C_9H_8NO_{3}^-][H^+]}{[HC_9H_8NO_{3}]}$

$K_a=\frac{x\times x}{(c -x)}$

$2.399\times 10^{-4}=\frac{x\times x}{(0.140 -x)}$

Solving for x:

x = 0.005677 M

$[H^+]=x = 0.005677 M$

The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[0.005677 M]=2.246\approx 2.2$

The pH in 0.140 M hippuric acid solution is 2.2.

6 0

4 years ago

___ V2O5 + ___ CaS ___ CaO + ___ V2S5

den301095 [7]

Answer:

Explanation:

5 0

3 years ago

How do we know the car has moved

Anuta_ua [19.1K]

Answer:

We need more information to solve this question. Take a picture of the problem and post it and I will try to help.

Explanation:

8 0

3 years ago

A sample of brass is put into a calorimeter (see sketch at right) that contains of water. The brass sample starts off at and the

Ierofanga [76]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The specific heat is $c_b = 0.402 J / g \cdot ^oC$

Explanation:

From the question we are told that

The mass of the sample is $m = 54.4 \ g$

The mass of the water is $m_w = 150.0 \ g$

The initial temperature of the sample is $T_i = 95.1 ^oC$

The initial temperature of the water is $T_{w_i} = 15^oC$

The final temperature of the water is $T = 17.6 ^oC$

Note the final temperature of water is equal to the final temperature of brass sample

The pressure is $P =1 \ atm$

Generally for according to the law of energy conservation

The heat lost by sample = The heat gain by water

The heat lost by brass sample is mathematically evaluated as

$H_L = m * c_b * [T_i - T]$

Where $c_b$ is the specific neat of the brass sample

The heat gained by water is mathematically evaluated as

$H_g = m_w *c_w * [T_w - T ]$

where $c_w$ is the specific heat of water which has a constant value of

$c_w = 4.186 joule/gram$

$H_L = H_g \ \equiv m* c_b * [T_i -T] = m_w * c_w * [T - T_w]$

substituting values

$52.4 * c_b * [95.1 - 17.6] = 150 * 4.186 * [ 17.6 - 15.0]$

$c_b = 0.402 J / g \cdot ^oC$

4 0

4 years ago